Mixing
Canonical equations from action integral
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I'm going through Lanczos Variational Principles of Classical Mechanics and on page 169 it says that we can form the action integral
$$A=int_t_1^t_2 left[sum p_idot q_i-H(q_1,...,q_n;p_1,...,p_n;t)right]dt$$
from which we can get do the variation to get
$$delta A=0=fracdp_idt+fracpartial Hpartial q_i$$
$$delta A=0=-dot q_i + fracpartial Hpartial p_i$$
Now this is straightforward for the $delta p$ variation,
$$p_ideltadot q_i + dot q_idelta p_i -fracpartial Hpartial q_idelta q_i - fracpartial Hpartial p_idelta p_i - fracpartial Hpartial t delta t = 0$$
$$rightarrow dot q_idelta p_i - fracpartial Hpartial p_idelta p_i = 0$$
but for the $delta q$ variation I'm finding I need to do something tricky looking and potentially incorrect:
$$p_ideltadot q_i + dot q_idelta p_i -fracpartial Hpartial q_idelta q_i - fracpartial Hpartial p_idelta p_i - fracpartial Hpartial t delta t = 0$$
$$rightarrow delta(fracddt q_i) p_i -fracpartial Hpartial q_idelta q_i = 0$$
and using the fact that d and $delta$ commute
$$rightarrow fracddtdelta(q_i) p_i -fracpartial Hpartial q_idelta q_i = 0$$
$$rightarrow fracddtp_i delta q_i -fracpartial Hpartial q_idelta q_i = 0$$
I can recover the equation as given, except I'm off by a minus sign! So, I'm wondering if the trick I used is valid, and why I'm getting a minus sign off the equation. Any help would be greatly appreciated!
calculus-of-variations
asked Jul 15 at 23:36
DS08
1078
add a comment |Â
up vote
0
down vote
favorite
I'm going through Lanczos Variational Principles of Classical Mechanics and on page 169 it says that we can form the action integral
$$A=int_t_1^t_2 left[sum p_idot q_i-H(q_1,...,q_n;p_1,...,p_n;t)right]dt$$
from which we can get do the variation to get
$$delta A=0=fracdp_idt+fracpartial Hpartial q_i$$
$$delta A=0=-dot q_i + fracpartial Hpartial p_i$$
Now this is straightforward for the $delta p$ variation,
$$p_ideltadot q_i + dot q_idelta p_i -fracpartial Hpartial q_idelta q_i - fracpartial Hpartial p_idelta p_i - fracpartial Hpartial t delta t = 0$$
$$rightarrow dot q_idelta p_i - fracpartial Hpartial p_idelta p_i = 0$$
but for the $delta q$ variation I'm finding I need to do something tricky looking and potentially incorrect:
$$p_ideltadot q_i + dot q_idelta p_i -fracpartial Hpartial q_idelta q_i - fracpartial Hpartial p_idelta p_i - fracpartial Hpartial t delta t = 0$$
$$rightarrow delta(fracddt q_i) p_i -fracpartial Hpartial q_idelta q_i = 0$$
and using the fact that d and $delta$ commute
$$rightarrow fracddtdelta(q_i) p_i -fracpartial Hpartial q_idelta q_i = 0$$
$$rightarrow fracddtp_i delta q_i -fracpartial Hpartial q_idelta q_i = 0$$
I can recover the equation as given, except I'm off by a minus sign! So, I'm wondering if the trick I used is valid, and why I'm getting a minus sign off the equation. Any help would be greatly appreciated!
calculus-of-variations
asked Jul 15 at 23:36
DS08
1078
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm going through Lanczos Variational Principles of Classical Mechanics and on page 169 it says that we can form the action integral
$$A=int_t_1^t_2 left[sum p_idot q_i-H(q_1,...,q_n;p_1,...,p_n;t)right]dt$$
from which we can get do the variation to get
$$delta A=0=fracdp_idt+fracpartial Hpartial q_i$$
$$delta A=0=-dot q_i + fracpartial Hpartial p_i$$
Now this is straightforward for the $delta p$ variation,
$$p_ideltadot q_i + dot q_idelta p_i -fracpartial Hpartial q_idelta q_i - fracpartial Hpartial p_idelta p_i - fracpartial Hpartial t delta t = 0$$
$$rightarrow dot q_idelta p_i - fracpartial Hpartial p_idelta p_i = 0$$
but for the $delta q$ variation I'm finding I need to do something tricky looking and potentially incorrect:
$$p_ideltadot q_i + dot q_idelta p_i -fracpartial Hpartial q_idelta q_i - fracpartial Hpartial p_idelta p_i - fracpartial Hpartial t delta t = 0$$
$$rightarrow delta(fracddt q_i) p_i -fracpartial Hpartial q_idelta q_i = 0$$
and using the fact that d and $delta$ commute
$$rightarrow fracddtdelta(q_i) p_i -fracpartial Hpartial q_idelta q_i = 0$$
$$rightarrow fracddtp_i delta q_i -fracpartial Hpartial q_idelta q_i = 0$$
I can recover the equation as given, except I'm off by a minus sign! So, I'm wondering if the trick I used is valid, and why I'm getting a minus sign off the equation. Any help would be greatly appreciated!
calculus-of-variations
asked Jul 15 at 23:36
DS08
1078
I'm going through Lanczos Variational Principles of Classical Mechanics and on page 169 it says that we can form the action integral
$$A=int_t_1^t_2 left[sum p_idot q_i-H(q_1,...,q_n;p_1,...,p_n;t)right]dt$$
from which we can get do the variation to get
$$delta A=0=fracdp_idt+fracpartial Hpartial q_i$$
$$delta A=0=-dot q_i + fracpartial Hpartial p_i$$
Now this is straightforward for the $delta p$ variation,
$$p_ideltadot q_i + dot q_idelta p_i -fracpartial Hpartial q_idelta q_i - fracpartial Hpartial p_idelta p_i - fracpartial Hpartial t delta t = 0$$
$$rightarrow dot q_idelta p_i - fracpartial Hpartial p_idelta p_i = 0$$
but for the $delta q$ variation I'm finding I need to do something tricky looking and potentially incorrect:
$$p_ideltadot q_i + dot q_idelta p_i -fracpartial Hpartial q_idelta q_i - fracpartial Hpartial p_idelta p_i - fracpartial Hpartial t delta t = 0$$
$$rightarrow delta(fracddt q_i) p_i -fracpartial Hpartial q_idelta q_i = 0$$
and using the fact that d and $delta$ commute
$$rightarrow fracddtdelta(q_i) p_i -fracpartial Hpartial q_idelta q_i = 0$$
$$rightarrow fracddtp_i delta q_i -fracpartial Hpartial q_idelta q_i = 0$$
I can recover the equation as given, except I'm off by a minus sign! So, I'm wondering if the trick I used is valid, and why I'm getting a minus sign off the equation. Any help would be greatly appreciated!
calculus-of-variations
asked Jul 15 at 23:36
DS08
1078
asked Jul 15 at 23:36
DS08
1078
asked Jul 15 at 23:36
DS08
1078
asked Jul 15 at 23:36
DS08
1078
1078
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
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up vote
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Your error is when going from $delta(fracddt q_i) p_i$ to $fracddtp_i delta q_i$. In doing so you must change sign:
$$
delta left( fracddt q_i right) p_i
= fracddt(delta q_i) p_i
= fracddt(delta q_i , p_i) - delta q_i , fracddtp_i
$$
Now, $fracddt(delta q_i , p_i)$ vanishes on integration if $delta q_i$ is taken to vanish at the end points, so we end with $- delta q_i , fracddtp_i.$
answered Jul 16 at 4:46
md2perpe
5,95511022
But what's the reasoning for changing the sign?
– DS08
Jul 17 at 16:04
It's a partial integration: $$int_t_1^t_2 fracd , delta q_idt p_i , dt = left[ delta q_i , p_i right]_t_1^t_2 - int_t_1^t_2 delta q_i fracd p_idt , dt$$
– md2perpe
Jul 17 at 16:09
add a comment |Â
up vote
0
down vote
You are treating $delta$ like a differential. $delta$ is a variation, not a differential. I feel like the only way I can tell you what went wrong in your manipulations, is to show you the correct way of dealing with a variation.
Consider an functional , which feeds one smooth vector function $mathbff$ and spits out an integral (which is called the action)
$$
S[mathbff] = int_a^b F(mathbff(t),dotmathbff(t),t) dt
$$
here $mathbff$ furthermore needs to satisfy $mathbff(a)=mathbfx$, $mathbff(b)=mathbfy$ with $mathbfx,mathbfy$ fixed and given. The function $F$ is say another smooth function.
What your book is trying to tell you is the function $mathbff$ making the action extremum, satisfies those equations. This is what one does. Let $mathbfg$ be the (unknown) extremum solution, $epsilon >0$ a small number and $mathbfh$ a function with $mathbfh(a)=mathbfh(b)=0$. This is a perturbation of the extremum solution. To put things in perspective, by $delta mathbff$ we mean $mathbff-mathbfg=epsilon mathbfh$. This is what a variation means.
Now
$$
F(mathbff(t),dotmathbff(t),t)-
F(mathbfg(t),dotmathbfg(t),t)
=
F(mathbfg+epsilonmathbfh,dotmathbfg+epsilon dotmathbfh,t)
-
F(mathbfg,dotmathbfg,t)
=
epsilonleft[
fracpartial Fpartial mathbffcdot mathbfh+
fracpartial Fpartial dotmathbffcdot dotmathbfh
right]
$$
the equality being up to $epsilon^2$ order. Therefore
$$
delta S := S[mathbff]-
S[mathbfg]=epsilon int_a^b
left[
fracpartial Fpartial mathbffcdot mathbfh+
fracpartial Fpartial dotmathbffcdot dotmathbfh
right]dt
$$
Doing an integration by parts over the second summand (and using the fact that $mathbfh(a)=mathbfh(b)=0$
$$
delta S =epsilon int_a^b
left[
fracpartial Fpartial mathbffcolorred-fracddt
fracpartial Fpartial dotmathbff
right]cdot mathbfh dt
$$
Now if $S$ is extremum at $mathbfg$, then $delta S =0$ for all choices of $mathbfh$. This forces
$$
boxed
fracpartial Fpartial mathbff-fracddt
fracpartial Fpartial dotmathbff=0
$$
which is called the Euler-Lagrange equation. You can either use, the ideas in this derivation to find Hamiltonian equations of motion from scratch, or you can just use the Euler-Lagrange equation to your action. I hope by now it is clear how unsafe what you were doing originally was.
answered Jul 16 at 2:37
Hamed
4,401421
Variations can be treated as differentials in the function space.
– md2perpe
Jul 17 at 16:13
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your error is when going from $delta(fracddt q_i) p_i$ to $fracddtp_i delta q_i$. In doing so you must change sign:
$$
delta left( fracddt q_i right) p_i
= fracddt(delta q_i) p_i
= fracddt(delta q_i , p_i) - delta q_i , fracddtp_i
$$
Now, $fracddt(delta q_i , p_i)$ vanishes on integration if $delta q_i$ is taken to vanish at the end points, so we end with $- delta q_i , fracddtp_i.$
answered Jul 16 at 4:46
md2perpe
5,95511022
But what's the reasoning for changing the sign?
– DS08
Jul 17 at 16:04
It's a partial integration: $$int_t_1^t_2 fracd , delta q_idt p_i , dt = left[ delta q_i , p_i right]_t_1^t_2 - int_t_1^t_2 delta q_i fracd p_idt , dt$$
– md2perpe
Jul 17 at 16:09
add a comment |Â
up vote
1
down vote
accepted
Your error is when going from $delta(fracddt q_i) p_i$ to $fracddtp_i delta q_i$. In doing so you must change sign:
$$
delta left( fracddt q_i right) p_i
= fracddt(delta q_i) p_i
= fracddt(delta q_i , p_i) - delta q_i , fracddtp_i
$$
Now, $fracddt(delta q_i , p_i)$ vanishes on integration if $delta q_i$ is taken to vanish at the end points, so we end with $- delta q_i , fracddtp_i.$
answered Jul 16 at 4:46
md2perpe
5,95511022
But what's the reasoning for changing the sign?
– DS08
Jul 17 at 16:04
It's a partial integration: $$int_t_1^t_2 fracd , delta q_idt p_i , dt = left[ delta q_i , p_i right]_t_1^t_2 - int_t_1^t_2 delta q_i fracd p_idt , dt$$
– md2perpe
Jul 17 at 16:09
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your error is when going from $delta(fracddt q_i) p_i$ to $fracddtp_i delta q_i$. In doing so you must change sign:
$$
delta left( fracddt q_i right) p_i
= fracddt(delta q_i) p_i
= fracddt(delta q_i , p_i) - delta q_i , fracddtp_i
$$
Now, $fracddt(delta q_i , p_i)$ vanishes on integration if $delta q_i$ is taken to vanish at the end points, so we end with $- delta q_i , fracddtp_i.$
answered Jul 16 at 4:46
md2perpe
5,95511022
Your error is when going from $delta(fracddt q_i) p_i$ to $fracddtp_i delta q_i$. In doing so you must change sign:
$$
delta left( fracddt q_i right) p_i
= fracddt(delta q_i) p_i
= fracddt(delta q_i , p_i) - delta q_i , fracddtp_i
$$
Now, $fracddt(delta q_i , p_i)$ vanishes on integration if $delta q_i$ is taken to vanish at the end points, so we end with $- delta q_i , fracddtp_i.$
answered Jul 16 at 4:46
md2perpe
5,95511022
answered Jul 16 at 4:46
md2perpe
5,95511022
answered Jul 16 at 4:46
md2perpe
5,95511022
answered Jul 16 at 4:46
md2perpe
5,95511022
5,95511022
But what's the reasoning for changing the sign?
– DS08
Jul 17 at 16:04
It's a partial integration: $$int_t_1^t_2 fracd , delta q_idt p_i , dt = left[ delta q_i , p_i right]_t_1^t_2 - int_t_1^t_2 delta q_i fracd p_idt , dt$$
– md2perpe
Jul 17 at 16:09
add a comment |Â
But what's the reasoning for changing the sign?
– DS08
Jul 17 at 16:04
It's a partial integration: $$int_t_1^t_2 fracd , delta q_idt p_i , dt = left[ delta q_i , p_i right]_t_1^t_2 - int_t_1^t_2 delta q_i fracd p_idt , dt$$
– md2perpe
Jul 17 at 16:09
But what's the reasoning for changing the sign?
– DS08
Jul 17 at 16:04
But what's the reasoning for changing the sign?
– DS08
Jul 17 at 16:04
It's a partial integration: $$int_t_1^t_2 fracd , delta q_idt p_i , dt = left[ delta q_i , p_i right]_t_1^t_2 - int_t_1^t_2 delta q_i fracd p_idt , dt$$
– md2perpe
Jul 17 at 16:09
It's a partial integration: $$int_t_1^t_2 fracd , delta q_idt p_i , dt = left[ delta q_i , p_i right]_t_1^t_2 - int_t_1^t_2 delta q_i fracd p_idt , dt$$
– md2perpe
Jul 17 at 16:09
add a comment |Â
up vote
0
down vote
You are treating $delta$ like a differential. $delta$ is a variation, not a differential. I feel like the only way I can tell you what went wrong in your manipulations, is to show you the correct way of dealing with a variation.
Consider an functional , which feeds one smooth vector function $mathbff$ and spits out an integral (which is called the action)
$$
S[mathbff] = int_a^b F(mathbff(t),dotmathbff(t),t) dt
$$
here $mathbff$ furthermore needs to satisfy $mathbff(a)=mathbfx$, $mathbff(b)=mathbfy$ with $mathbfx,mathbfy$ fixed and given. The function $F$ is say another smooth function.
What your book is trying to tell you is the function $mathbff$ making the action extremum, satisfies those equations. This is what one does. Let $mathbfg$ be the (unknown) extremum solution, $epsilon >0$ a small number and $mathbfh$ a function with $mathbfh(a)=mathbfh(b)=0$. This is a perturbation of the extremum solution. To put things in perspective, by $delta mathbff$ we mean $mathbff-mathbfg=epsilon mathbfh$. This is what a variation means.
Now
$$
F(mathbff(t),dotmathbff(t),t)-
F(mathbfg(t),dotmathbfg(t),t)
=
F(mathbfg+epsilonmathbfh,dotmathbfg+epsilon dotmathbfh,t)
-
F(mathbfg,dotmathbfg,t)
=
epsilonleft[
fracpartial Fpartial mathbffcdot mathbfh+
fracpartial Fpartial dotmathbffcdot dotmathbfh
right]
$$
the equality being up to $epsilon^2$ order. Therefore
$$
delta S := S[mathbff]-
S[mathbfg]=epsilon int_a^b
left[
fracpartial Fpartial mathbffcdot mathbfh+
fracpartial Fpartial dotmathbffcdot dotmathbfh
right]dt
$$
Doing an integration by parts over the second summand (and using the fact that $mathbfh(a)=mathbfh(b)=0$
$$
delta S =epsilon int_a^b
left[
fracpartial Fpartial mathbffcolorred-fracddt
fracpartial Fpartial dotmathbff
right]cdot mathbfh dt
$$
Now if $S$ is extremum at $mathbfg$, then $delta S =0$ for all choices of $mathbfh$. This forces
$$
boxed
fracpartial Fpartial mathbff-fracddt
fracpartial Fpartial dotmathbff=0
$$
which is called the Euler-Lagrange equation. You can either use, the ideas in this derivation to find Hamiltonian equations of motion from scratch, or you can just use the Euler-Lagrange equation to your action. I hope by now it is clear how unsafe what you were doing originally was.
answered Jul 16 at 2:37
Hamed
4,401421
Variations can be treated as differentials in the function space.
– md2perpe
Jul 17 at 16:13
add a comment |Â
up vote
0
down vote
You are treating $delta$ like a differential. $delta$ is a variation, not a differential. I feel like the only way I can tell you what went wrong in your manipulations, is to show you the correct way of dealing with a variation.
Consider an functional , which feeds one smooth vector function $mathbff$ and spits out an integral (which is called the action)
$$
S[mathbff] = int_a^b F(mathbff(t),dotmathbff(t),t) dt
$$
here $mathbff$ furthermore needs to satisfy $mathbff(a)=mathbfx$, $mathbff(b)=mathbfy$ with $mathbfx,mathbfy$ fixed and given. The function $F$ is say another smooth function.
What your book is trying to tell you is the function $mathbff$ making the action extremum, satisfies those equations. This is what one does. Let $mathbfg$ be the (unknown) extremum solution, $epsilon >0$ a small number and $mathbfh$ a function with $mathbfh(a)=mathbfh(b)=0$. This is a perturbation of the extremum solution. To put things in perspective, by $delta mathbff$ we mean $mathbff-mathbfg=epsilon mathbfh$. This is what a variation means.
Now
$$
F(mathbff(t),dotmathbff(t),t)-
F(mathbfg(t),dotmathbfg(t),t)
=
F(mathbfg+epsilonmathbfh,dotmathbfg+epsilon dotmathbfh,t)
-
F(mathbfg,dotmathbfg,t)
=
epsilonleft[
fracpartial Fpartial mathbffcdot mathbfh+
fracpartial Fpartial dotmathbffcdot dotmathbfh
right]
$$
the equality being up to $epsilon^2$ order. Therefore
$$
delta S := S[mathbff]-
S[mathbfg]=epsilon int_a^b
left[
fracpartial Fpartial mathbffcdot mathbfh+
fracpartial Fpartial dotmathbffcdot dotmathbfh
right]dt
$$
Doing an integration by parts over the second summand (and using the fact that $mathbfh(a)=mathbfh(b)=0$
$$
delta S =epsilon int_a^b
left[
fracpartial Fpartial mathbffcolorred-fracddt
fracpartial Fpartial dotmathbff
right]cdot mathbfh dt
$$
Now if $S$ is extremum at $mathbfg$, then $delta S =0$ for all choices of $mathbfh$. This forces
$$
boxed
fracpartial Fpartial mathbff-fracddt
fracpartial Fpartial dotmathbff=0
$$
which is called the Euler-Lagrange equation. You can either use, the ideas in this derivation to find Hamiltonian equations of motion from scratch, or you can just use the Euler-Lagrange equation to your action. I hope by now it is clear how unsafe what you were doing originally was.
answered Jul 16 at 2:37
Hamed
4,401421
Variations can be treated as differentials in the function space.
– md2perpe
Jul 17 at 16:13
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You are treating $delta$ like a differential. $delta$ is a variation, not a differential. I feel like the only way I can tell you what went wrong in your manipulations, is to show you the correct way of dealing with a variation.
Consider an functional , which feeds one smooth vector function $mathbff$ and spits out an integral (which is called the action)
$$
S[mathbff] = int_a^b F(mathbff(t),dotmathbff(t),t) dt
$$
here $mathbff$ furthermore needs to satisfy $mathbff(a)=mathbfx$, $mathbff(b)=mathbfy$ with $mathbfx,mathbfy$ fixed and given. The function $F$ is say another smooth function.
What your book is trying to tell you is the function $mathbff$ making the action extremum, satisfies those equations. This is what one does. Let $mathbfg$ be the (unknown) extremum solution, $epsilon >0$ a small number and $mathbfh$ a function with $mathbfh(a)=mathbfh(b)=0$. This is a perturbation of the extremum solution. To put things in perspective, by $delta mathbff$ we mean $mathbff-mathbfg=epsilon mathbfh$. This is what a variation means.
Now
$$
F(mathbff(t),dotmathbff(t),t)-
F(mathbfg(t),dotmathbfg(t),t)
=
F(mathbfg+epsilonmathbfh,dotmathbfg+epsilon dotmathbfh,t)
-
F(mathbfg,dotmathbfg,t)
=
epsilonleft[
fracpartial Fpartial mathbffcdot mathbfh+
fracpartial Fpartial dotmathbffcdot dotmathbfh
right]
$$
the equality being up to $epsilon^2$ order. Therefore
$$
delta S := S[mathbff]-
S[mathbfg]=epsilon int_a^b
left[
fracpartial Fpartial mathbffcdot mathbfh+
fracpartial Fpartial dotmathbffcdot dotmathbfh
right]dt
$$
Doing an integration by parts over the second summand (and using the fact that $mathbfh(a)=mathbfh(b)=0$
$$
delta S =epsilon int_a^b
left[
fracpartial Fpartial mathbffcolorred-fracddt
fracpartial Fpartial dotmathbff
right]cdot mathbfh dt
$$
Now if $S$ is extremum at $mathbfg$, then $delta S =0$ for all choices of $mathbfh$. This forces
$$
boxed
fracpartial Fpartial mathbff-fracddt
fracpartial Fpartial dotmathbff=0
$$
which is called the Euler-Lagrange equation. You can either use, the ideas in this derivation to find Hamiltonian equations of motion from scratch, or you can just use the Euler-Lagrange equation to your action. I hope by now it is clear how unsafe what you were doing originally was.
answered Jul 16 at 2:37
Hamed
4,401421
You are treating $delta$ like a differential. $delta$ is a variation, not a differential. I feel like the only way I can tell you what went wrong in your manipulations, is to show you the correct way of dealing with a variation.
Consider an functional , which feeds one smooth vector function $mathbff$ and spits out an integral (which is called the action)
$$
S[mathbff] = int_a^b F(mathbff(t),dotmathbff(t),t) dt
$$
here $mathbff$ furthermore needs to satisfy $mathbff(a)=mathbfx$, $mathbff(b)=mathbfy$ with $mathbfx,mathbfy$ fixed and given. The function $F$ is say another smooth function.
What your book is trying to tell you is the function $mathbff$ making the action extremum, satisfies those equations. This is what one does. Let $mathbfg$ be the (unknown) extremum solution, $epsilon >0$ a small number and $mathbfh$ a function with $mathbfh(a)=mathbfh(b)=0$. This is a perturbation of the extremum solution. To put things in perspective, by $delta mathbff$ we mean $mathbff-mathbfg=epsilon mathbfh$. This is what a variation means.
Now
$$
F(mathbff(t),dotmathbff(t),t)-
F(mathbfg(t),dotmathbfg(t),t)
=
F(mathbfg+epsilonmathbfh,dotmathbfg+epsilon dotmathbfh,t)
-
F(mathbfg,dotmathbfg,t)
=
epsilonleft[
fracpartial Fpartial mathbffcdot mathbfh+
fracpartial Fpartial dotmathbffcdot dotmathbfh
right]
$$
the equality being up to $epsilon^2$ order. Therefore
$$
delta S := S[mathbff]-
S[mathbfg]=epsilon int_a^b
left[
fracpartial Fpartial mathbffcdot mathbfh+
fracpartial Fpartial dotmathbffcdot dotmathbfh
right]dt
$$
Doing an integration by parts over the second summand (and using the fact that $mathbfh(a)=mathbfh(b)=0$
$$
delta S =epsilon int_a^b
left[
fracpartial Fpartial mathbffcolorred-fracddt
fracpartial Fpartial dotmathbff
right]cdot mathbfh dt
$$
Now if $S$ is extremum at $mathbfg$, then $delta S =0$ for all choices of $mathbfh$. This forces
$$
boxed
fracpartial Fpartial mathbff-fracddt
fracpartial Fpartial dotmathbff=0
$$
which is called the Euler-Lagrange equation. You can either use, the ideas in this derivation to find Hamiltonian equations of motion from scratch, or you can just use the Euler-Lagrange equation to your action. I hope by now it is clear how unsafe what you were doing originally was.
answered Jul 16 at 2:37
Hamed
4,401421
answered Jul 16 at 2:37
Hamed
4,401421
answered Jul 16 at 2:37
Hamed
4,401421
answered Jul 16 at 2:37
Hamed
4,401421
4,401421
Variations can be treated as differentials in the function space.
– md2perpe
Jul 17 at 16:13
add a comment |Â
Variations can be treated as differentials in the function space.
– md2perpe
Jul 17 at 16:13
Variations can be treated as differentials in the function space.
– md2perpe
Jul 17 at 16:13
Variations can be treated as differentials in the function space.
– md2perpe
Jul 17 at 16:13
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